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DENIUS [597]
3 years ago
10

Jay and Talia are hiking different trails. In the middle of their hikes, jay is 20 feet below sea level and Talia is the same nu

mber of feet above sea level. Graph Jays elevation at -20 on the number line. Then graph Tailias elevation on the number line. Click on the number line to identify the locations of the points.
Mathematics
1 answer:
topjm [15]3 years ago
5 0

Answer:

you can do this i believe in you

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Where do i graph it
Anuta_ua [19.1K]
(-5,-10)E
(-5,-3)F
(-3,-10)G
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3 years ago
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6x-58+2x+4=5x<br>find the value of x
mariarad [96]
You pass the x in one side and numbers to the other side
8x-5x=58-4
add and subtract
3x=54
divide by 3
x=18
7 0
3 years ago
What is “2 1/2 x 3 1/3”
lesya692 [45]

Answer:

The answer is 8.3 bar notation.

Step-by-step explanation:

2 1/2 into improper is 5/2

3 1/3 into improper is 10/3

Now Multiply

5/2 x 10/3 is 300/36

Now Reduce.

25/3 = 8.3 bar notation

Mark Brainliest!

3 0
2 years ago
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Sketch the domain D bounded by y = x^2, y = (1/2)x^2, and y=6x. Use a change of variables with the map x = uv, y = u^2 (for u ?
cluponka [151]

Under the given transformation, the Jacobian and its determinant are

\begin{cases}x=uv\\y=u^2\end{cases}\implies J=\begin{bmatrix}v&u\\2u&0\end{bmatrix}\implies|\det J|=2u^2

so that

\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\iint_{D'}\frac{2u^2}{u^2}\,\mathrm du\,\mathrm dv=2\iint_{D'}\mathrm du\,\mathrm dv

where D' is the region D transformed into the u-v plane. The remaining integral is the twice the area of D'.

Now, the integral over D is

\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\left\{\int_0^6\int_{x^2/2}^{x^2}+\int_6^{12}\int_{x^2/2}^{6x}\right\}\frac{\mathrm dx\,\mathrm dy}y

but through the given transformation, the boundary of D' is the set of equations,

\begin{array}{l}y=x^2\implies u^2=u^2v^2\implies v^2=1\implies v=\pm1\\y=\frac{x^2}2\implies u^2=\frac{u^2v^2}2\implies v^2=2\implies v=\pm\sqrt2\\y=6x\implies u^2=6uv\implies u=6v\end{array}

We require that u>0, and the last equation tells us that we would also need v>0. This means 1\le v\le\sqrt2 and 0, so that the integral over D' is

\displaystyle2\iint_{D'}\mathrm du\,\mathrm dv=2\int_1^{\sqrt2}\int_0^{6v}\mathrm du\,\mathrm dv=\boxed6

4 0
3 years ago
Pls help me out i’ll give brainliest to best answer
Vedmedyk [2.9K]

Answer:

5. ABC and XYZ

Step-by-step explanation:

matching angle values

5 0
2 years ago
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